# Complex Numbers with Vector Addition

Complex Numbers with Vector Addition are obtained using the sides of triangles; the sum of sides of two sides of a triangle is greater than the other side.

### Worked on Examples of Complex Numbers with Vector Addition

Suppose $\displaystyle 0 \lt \alpha, \ \beta \lt \frac{\pi}{2}$ and define complex numbers $z_n$ by
$$z_n = \cos(\alpha + n \beta) + i \sin(\alpha + n \beta)$$
for $n = 0, \ 1, \ 2, \ 3, \ 4$. the points $P_0, \ P_1, \ P_2$ and $P_3$ are the points in the Argand diagram that correspond to the complex numbers $z_0, \ z_0+z_1, \ z_0+z_1+z_2$ and $z_0+z_1+z_2+z_3$ respectively. The angles $\theta_0, \ \theta_1$ and $\theta_2$ are the external angles at $P_0, \ P_1$ and $P_2$.

(a)    Using vector addition, explain why $\theta_0 = \theta_1 = \theta_2 = \beta$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} z_0 &= \cos(\alpha) + i \sin(\alpha) \color{red} \cdots (0) \\ z_1 &= \cos(\alpha + \beta) + i \sin(\alpha + \beta) \color{red} \cdots (1) \\ z_2 &= \cos(\alpha + 2 \beta) + i \sin(\alpha + 2 \beta) \color{red} \cdots (2) \\ z_3 &= \cos(\alpha + 3 \beta) + i \sin(\alpha + 3 \beta) \color{red} \cdots (3) \\ z_4 &= \cos(\alpha + 4 \beta) + i \sin(\alpha + 4 \beta) \color{red} \cdots (4) \\ \overrightarrow{z_0} &= \overrightarrow{OP_0} \\ \angle P_0 O x &= \alpha &\color{red} \text{by } (0) \\ \overrightarrow{OP_1} &= \overrightarrow{OP_0} + \overrightarrow{P_0 P_1} \\ \overrightarrow{z_0 + z_1} &= \overrightarrow{z_0} + \overrightarrow{z_1} \\ \overrightarrow{z_0 + z_1} &= \overrightarrow{OP_0} + \overrightarrow{P_0 P_1} \\ \overrightarrow{z_1} &= \overrightarrow{P_0 P_1} \\ z_0 &= \cos(\alpha + \theta_0) + i \sin(\alpha + \theta_0) \\ \theta_0 &= \beta &\color{red} \text{by } (1) \\ z_1 &= \cos(\alpha + \theta_0 + \theta_1) + i \sin(\alpha + \theta_0 + \theta_1) \\ &= \cos(\alpha + \beta + \theta_1) + i \sin(\alpha + \beta + \theta_1) \\ \theta_1 &= \beta &\color{red} \text{by } (2) \\ z_2 &= \cos(\alpha + \theta_0 + \theta_1 + \theta_2) + i \sin(\alpha + \theta_0 + \theta_1 + \theta_2) \\ &= \cos(\alpha + \beta + \beta + \theta_2) + i \sin(\alpha + \beta + \beta + \theta_2) \\ \theta_2 &= \beta &\color{red} \text{by } (3) \\ \therefore \theta_0 &= \theta_1 = \theta_2 = \beta \\ \end{aligned} \\

(b)    Show that $\angle P_0O P_1 = \angle P_0 P_2 P_1$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \angle O P_0 P_1 &= \angle P_0 P_1 P_2 \color{red} \cdots (5) &\color{red} \theta_0 = \theta_1 = \beta \\ OP_0 &= P_0 P_1 = P_1 P_2 = 1 \color{red} \cdots (6) \\ \triangle O P_0 P_1 &\equiv \triangle P_0 P_1 P_2 &\color{red} \text{by } (5), (6) \\ \therefore \angle P_0O P_1 &= \angle P_0 P_2 P_1 &\color{red} \text{corresponding angles are equal} \end{aligned}

(c)    Explain why $O P_0 P_1 P_2$ is a cyclic quadrilateral.

\begin{aligned} \displaystyle\require{AMSsymbols} \require{color} \angle P_0O P_1 &= \angle P_0 P_2 P_1 &\color{red} \text{by } (b) \\ \therefore O P_0 P_1 P_2 &\text{ is a cyclic quadrilateral} &\color{red} \text{angles subtended by a chord} \end{aligned}

(d)    Show that $P_0 P_1 P_2 P_3$ is a cyclic quadrilateral.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \angle P_0 P_1 P_2 &= \angle P_1 P_2 P_3 \color{red} \cdots (7) &\color{red} \theta_1 = \theta_2 = \beta \\ P_0 P_1 &= P_1 P_2 = P_2 P_3 = 1 \color{red} \cdots (8) \\ \triangle P_0 P_1 P_2 &\equiv \triangle P_1 P_2 P_3 &\color{red} \text{by } (7), (8) \\ \angle P_1 P_0 P_2 &= \angle P_2 P_3 P_1 &\color{red} \text{corresponding angles are equal} \\ \therefore P_0 P_1 P_2 P_3 &\text{ is a cyclic quadrilateral} &\color{red} \text{angles subtended by a chord} \end{aligned}

(e)    Explain why $O P_0 P_1 P_2 P_3$ are concyclic.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} O P_0 P_1 P_2 &\text{ are concyclic } \color{red} \cdots (9) \\ P_0 P_1 P_2 P_3 &\text{ are concyclic } \color{red} \cdots (10) \\ \therefore O P_0 P_1 P_2 P_3 &\text{ are concyclic } &\color{red} \text{by } (9), (10) \end{aligned}

(f)    Find the value of $\beta$ if $z_0 + z_1 + z_2 +z_3 + z_4 = 0$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \overrightarrow{OP_4} &= \overrightarrow{z_0 + z_1 + z_2 +z_3 + z_4} \\ \overrightarrow{OP_4} &= 0 \\ \text{Thus } &P_4 \text{ is at } O. \\ P_0 P_1 P_2 P_3 P_4 &\text{ are concyclic}. \\ P_0 P_1 P_2 P_3 P_4 &\text{ is a regular pentagon}. &\color{red} \text{equal exterior angles and equal sides} \\ \theta_0 &= \theta_1 = \theta_2 = \theta_3 = \theta_4 = \beta &\color{red} \text{exterior angles} \\ \therefore \beta &= \frac{2 \pi}{5} &\color{red} \text{sum of exterior angles is } 2 \pi \end{aligned}